SOLUTION: For what value of k does the following system have infinitely many solutions?
kx + y + z=0
x + 2y + kz = 0
-X+3z=0
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Question 1198864: For what value of k does the following system have infinitely many solutions?
kx + y + z=0
x + 2y + kz = 0
-X+3z=0
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
Solve the 3rd equation for z and substitute in the other two equations.
x = 3z
3kz+y+z=0
3z+2y+kz=0
y+(3k+1)z=0
2y+(3+k)z=0
There are infinitely many solutions if the two equations are equivalent.
2y+(6k+2)z=0
2y+(3+k)z=0
6k+2=3+k
5k=1
k=1/5
ANSWER: k=1/5
Answer by ikleyn(52786) (Show Source): You can put this solution on YOUR website!
.
This system has infinitely many solutions if and only if the determinant of the
coefficient matrix is zero.
So, calculate the determinant of the matrix as a function of "k" and equate it to zero.
From this equation, find the value of "k".
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